PUBLICATIONS DU DÉPARTEMENT DE MATHÉMATIQUES DE LYON
W. B. VASANTHA KANDASAMY On K-Boolean Rings Publications du Département de Mathématiques de Lyon, 1992, fascicule 2A , p. 73-77
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques http://www.numdam.org/ On K -Boolean Rings
W.B.Vasantha Kandasamy Department of Mathematics, Indian Institute of Technology, Madras - 600 036, India.
In this note we define K-Boolean rings which are a
generalization of a Boolean ring and obtain conditions for a
ring to be K-Boolean in general and in particular for a group
ring RG to be K-Boolean, where R denotes a ring and G a group,
RG the group ring of G over R.
Definition 1. Let R be a ring with identity we say R is a 2k K-Boolean ring if x = x for every x£R and for some natural
number k. Remarko When k = 1 we trivially get R to be a Boolean ring.
Proposition 2. Every Boolean ring is a K-Boolean ring*
o Proof. Given R is a Boolean ring hence x = x for every x£R; 9k so clearly for every x £ R x = x. Hence R is K-Boolean.
Proposition 3. Every k-Boolean ring need not be a Boolean ring.
Proof. By an example. Take 2^ = to be the field of
3 G C 2 characteristic two and G=
2 1+9f l+92> 9+92> l+9+g Jj is 2-Boolean which is clearly not
Boolean as 1+g is in 'Z2G but (1+g) + 1+g. Hence the result.
Example. Let Z, = (0,1) be the field of characteristic two and
2 G is not n G =
1+gkZ^G but (1+g)2 = 0. In view of the above example we have the f ollowing. 73 Proposition 4« Let Z2 = (0,1) be a field of characteristic 2n two and G =
the group ring Z2G is not a n-Boolean ring for any n.
n 11 2 Proof. Clearly l+g £ Z2G with (1+g ) = 0; hence Z2G is not
n-Boolean for any n# The above result can be generalized to any commutative ring of characteristic two as follows.
Theorem 5# Let R be a commutative ring of characteristic two 2n and G =
Proof. As in the case of proposition 4 we have l+gn£RG with
(l+gn)2 = 0. hence the group ring RG is not a n-Boolean ring for any n*
Proposition 6o If R is a n-Boolean ring then R has no non zero nilpotent elements.
P roof. Obvious.
Theorem 7# Let Z2 = (O.l) be the field of characteristic two and G =
Proof. Take any a^^G, since We have for every g£G ; 2n+l 2(n+l) g = l , a = a for every a(rZ2G. Thus Z2G is (n+l)-Boolean. 2(n+l) Remark. If 2(n+l) 42 for Some s we wil1 not nave a = a* 5 6 By an example take G =
74 Theorem 8. Let R be a commutative ring of characteristic
two with no nontxivial nilpotents and in which every element 2(n+l) 2n+l T in R is of the form -y = y and G =
group ring RG is y-Boolean with r = n+1 and 2(n+l) = 2s(s>l).
Proof, Obvious.
Theorem 9« Let R be a n-Boolean ring then characteristic of
R is two.
Proof. Four possibilities arise (i) characteristic of R is odd
(ii) characteristic of R is even not-equal to 2. (iii)
characteristic of R is 2n and (iv) characteristic of R is zero.
Case (i)> Let characteristic of R be odd; say 2n+l. Since 2 lCr R we have an integer m£R with m <2n+l with m = 1. Thus
R is not n-Boolean for any integer n. Case (ii). Let characteristic of R be even say 2n, since l£ R 2 clearly 2n-l£- R but (2n-l) = 1 hence R is not n-Boolean for
any integer n.
Case (iii). Characteristic of R is 2n (n>l); 2n 2 Let m s -j- + 1 then m = 1 so R is not n-Boolean.
Case (iv). Let characteristic of R be zero; since 16 R
+ we have Z UZ~u£o^ C R. Clearly for no natural integer
+ n(-Z+UZ~U {0^ we have nr = n. Hence R is n-Boolean. Hence
the characteristic of R is two.
Theorem 10. Let RG be the group ring of a group G over the
ring R and G a group in which every element is of finite order.
RG is n-Boolean if and only if R is a n-Boolean ring of
characteristic two and G is a commutative group in which every
element is of odd order m with m+1 = 23 for some s>l.
75 Proof, Since 1£ G we have R.l = RCRG. Let us asame RG
to be n-Boolean this implies by Theorem 9 and definition
of a n-Boolean ring characteristic of RG is two and R is
commutative since RCRG and R is n-Boolean ring. G is a
commutative group since RG is commutative further every
element in G is of odd order for if an element g is of even
order we have g^l so (l+g+.. •+g2r""1)2 * 0 hence RG is not n-Boolean. Hence the claim.
Conversely if R is n-Boolean ring and G is a commutative group
in which every element is of odd order with m+1 = 2s, clearly
RG is n-Boolean as RG is commutative with characteristic of RG 2k to be two and every element some k as every element in G is of odd order. Hence the result. In the above theorem we must have m+1 = 2 for if m+1 4 2S for any s we will not have a2^*1^ a. By an example take G = = (i+g4)(i+g)6 = U+g4)U+g)4U+g)2 = U+g4)U+g4)U+g2) « 8 2 i S (l+g )(l+g ) 4 l+g* Hence we must have m+1 = 2 ; then only when we expand we will always have the power of that element to be exactly divided in twos till the end. Theorem 11. Let R be a finite commutative ring of characteristic two with no non zero nilpotent elements. Then R is a K-Boolean ring if and only if every element in R is of odd order m with m+1 = 2s. 76 Proof* One way is obvious, for if every element is of odd order m with (IÎH-JL) = 2 then clearly R is a K-Boolean ring. Conversely if R is a K-Boolean ring then we have 2k x s x for every xé R ; if K is not a odd number such that k+1 4 2s then we have (l+x)2k 4 1+x. Hence the theorem. Ref erence [l] Jacobson, N. Structure of Rings, A.M.S. (1956). 77